# Integration by Parts

## Related calculator: Integral (Antiderivative) Calculator with Steps

It is easy to compute the integral int e^xdx, but how to handle integrals like int xe^xdx?
In general, if you have under the integral sign a product of functions that can be easily integrated separately, you should use integration by parts.

Formula for integration by parts: int udv=uv-int vdu.

Proof

Using the product rule, we have that (f(x)g(x))'=f'(x)g(x)+f(x)g'(x).

Integrating both sides gives: int (f(x)g(x))'dx=int (f'(x)g(x)+f(x)g'(x))dx.

This can be rewritten as f(x)g(x)=int f'(x)g(x)dx+int f(x)g'(x)dx or int f(x)g'(x)dx=f(x)g(x)-int f'(x)g(x)dx.

If we take u=f(x) and v=g(x), we have that du=f'(x)dx and dv=g'(x)dx, and the above formula can be rewritten using the substitution rule as int udv=uv-int vdu.

As can be seen, integration by parts corresponds to the product rule (just like the substitution rule corresponds to the chain rule).

In fact, every differentiation rule has a corresponding integration rule, because these processes are the inverse of each other.

Example 1. Evaluate int xe^xdx.

Let u=x and dv=e^xdx. Then, du=(x)'dx=dx, and v=int e^xdx=e^x.

So, int xe^xdx=int overbrace (x)^u overbrace (e^xdx)^(dv)=overbrace (x)^u overbrace (e^x)^v-int overbrace (e^x)^v overbrace(dx)^(du)=xe^x-e^x+C.

Note that it is very important to choose appropriate u and v, because a wrong choice will only complicate the integral.

For example, assume that, instead of choosing u=x and dv=e^xdx in the above example, we choose u=e^x and v=xdx. Then, du=(e^x)'dx=e^xdx, and v=int xdx=x^2/2.

So, int xe^xdx=int overbrace (e^x)^u overbrace (xdx)^(dv)=overbrace (e^x)^u overbrace (1/2 x^2)^v-int overbrace (1/2 x^2)^v overbrace(e^xdx)^(du).

Although this equation is true, the integral int 1/2 x^2 e^xdx is more difficult to evaluate than the integral we started with.

Example 2. Evaluate int x^2 cos(x)dx.

Notice that x^2 becomes simpler when differentiated. Therefore, let u=x^2 and dv=cos(x)dx. Then, du=(x^2)'dx=2xdx, and v=int cos(x)dx=sin(x).

So, int x^2cos(x)dx=x^2sin(x)-int 2xsin(x)dx=x^2sin(x)-2int xsin(x)dx.

We've obtained a simpler integral, but still it is not obvious. Therefore, we apply integration by parts once more to the integral int xsin(x)dx:

let u=x and dv=sin(x)dx; then, du=(x)'dx=dx, and v=int sin(x)dx=-cos(x).

So, int overbrace (x)^u overbrace (sin(x)dx)^(dv)=overbrace (x)^u overbrace (-cos(x))^v-int overbrace (-cos(x))^(v) overbrace (dx)^(du)=-xcos(x)+int cos(x)dx=

=-xcos(x)+sin(x)+C.

Finally, int x^2 cos(x)dx=x^2sin(x)-2int xsin(x)dx=x^2sin(x)-2(-xcos(x)+sin(x)+C)=

=x^2sin(x)+2xcos(x)-2sin(x)+C_1 where C_1=-2C.

Example 2 shows that in some cases we need to apply integration by parts more than once.

Example 3. Evaluate int ln(x)dx.

Here, there is only one choice for u and v, namely u=ln(x) and dv=dx; so,du=(ln(x))'dx=1/x dx, and v=int dx=x.

Therefore, int overbrace (ln(x))^u overbrace (dx)^(dv)=overbrace (ln(x))^u overbrace (x)^v-int overbrace (x)^v overbrace (1/xdx)^(du)=

=xln(x)-int dx=xln(x)-x+C.

Example 4. Evaluate int e^xsin(x)dx.

Neither e^x nor sin(x) become simpler when differentiated, but we try choosing u=e^x and v=sin(x)dx.

Then, du=(e^x)'dx=e^xdx, and v=int sin(x)dx=-cos(x)dx.

Therefore, int e^x sin(x)dx=-e^xcos(x)-int -cos(x)e^xdx=-e^xcos(x)+int e^xcos(x)dx.

The integral obtained is not simpler than the original one, but it is not more difficult either. So, we apply integration by parts once more: let u=e^x and dv=cos(x)dx; then, du=(e^x)'dx=e^xdx, and v=int cos(x)dx=sin(x).

So, int e^xcos(x)dx=e^xsin(x)-int e^xsin(x)dx.

We've obtained the initial integral! It seems that we obtained nothing because we've arrived at int e^xsin(x)dx, which is where we had started.

However, since int e^x sin(x)dx=-e^xcos(x)+int e^xcos(x)dx, it can be stated that

int e^x sin(x)dx=-e^xcos(x)+(e^x sin(x)-int e^xsin(x)dx).

This can be regarded as an equation with the unknown variable int e^x sin(x)dx.

This equation can be rewritten as 2int e^xsin(x)dx=e^x(sin(x)-cos(x)).

Dividing by 2 and adding the constant of integration yields the final answer: int e^xsin(x)dx=1/2 e^x(sin(x)-cos(x))+C.

Example 5. Prove the reduction formula.int sin^n(x)dx=-1/n cos(x)sin^(n-1)(x)+(n-1)/n int sin^(n-2)(x)dx, where n>=2 is an integer.

Let u=sin^(n-1)(x) and dv=sin(x)dx; then,du=(sin^(n-1))'dx=(n-1)sin^(n-2)(x)(sin(x))'dx=(n-1)sin^(n-2)(x)cos(x)dx, and v=int sin(x)dx=-cos(x).

So, int sin^n(x)dx=int sin^(n-1)sin(x)dx=-cos(x)sin^(n-1)(x)-int cos(x)(n-1)sin^(n-2)(x)cos(x)dx=

=-cos(x)sin^(n-1)(x)+(n-1) int sin^(n-2)(x) cos^2(x)dx.

Since cos^2(x)=1-sin^2(x), it can be stated that(n-1)int sin^(n-2)(x)cos^2(x)dx=(n-1)int sin^(n-2)(x)(1-sin^2(x))dx=

=(n-1)int sin^(n-2)(x)dx-(n-1)int sin^(n-2)sin^2(x)dx=

=(n-1)int sin^(n-2)(x)dx-(n-1)int sin^n(x)dx.

Therefore,

int sin^n(x)dx=-cos(x)sin^(n-1)(x)+(n-1)int sin^(n-2)(x)dx-(n-1) int sin^n(x)dx.

This is an equation with an unknown variable int sin^n(x)dx.

It can be rewritten as n int sin^n(x)dx=-cos(x)sin^(n-1)(x)+(n-1)int sin^(n-2)(x)dx.

So, finally, we have that int sin^n(x)dx=-1/n cos(x)sin^(n-1)(x)+(n-1)/n int sin^(n-2)(x)dx.

The last question in this section is how to calculate definite integrals with the help of integration by parts. In fact, it is very easy: just combine integration by parts with the Newton-Leibniz formula: int_a^b udv=uv|_a^b-int_a^b vdu.

Example 6. Calculate int_0^1 tan^(-1)(x)dx.

Let u=tan^(-1)(x) and dv=dx; then, du=(tan^(-1)(x))'dx=1/(x^2+1)dx, and v=int dx=x.

So, int_0^1 tan^(-1)(x)dx=xtan^(-1)(x)|_0^1-int_0^1 x/(x^2+1)dx=

=(1*tan^(-1)(1)-0*tan^(-1)(0))-int_0^1 x/(x^2+1)dx=pi/4-int_0^1 x/(x^2+1)dx.

To calculate int_0^1 x/(x^2+1)dx, we use the substitution rule: let t=x^2+1; then,dt=(x^2+1)'dx=2xdx, or xdx=1/2 dt.

Since x is changing from 0 to 1, it can be stated that t is changing from 2*0=0 to 2*1=2.

So, int_0^1 x/(x^2+1)dx=int_1^2 1/t 1/2 dt=1/2 [ln|t|]_1^2=1/2(ln(2)-ln(1))=

=1/2(ln(2)-0)=1/2 ln(2).

Finally, we can see that int_0^1 tan^(-1)(x)=pi/4-int_0^1 x/(x^2+1)dx=pi/4-1/2ln(2).